For a uniform probability model, we consider all possible pairs. Total ways to select 2 balls from 12: C(12,2) = 66. Ways to select 2 red: C(3,2) = 3. Ways to select 2 blue: C(4,2) = 6. Ways to select 2 green: C(5,2) = 10. Total favorable outcomes: 3 + 6 + 10 = 19. Probability = 19/66. Choice B incorrectly adds one extra outcome, C assumes equal colors, D includes mixed color pairs.

When you encounter a probability question involving specific outcomes, start by identifying exactly what event needs to happen and how many ways it can occur.

Here, the player is on space 47 and needs to land exactly on space 50. To move from space 47 to space 50, they need to advance exactly 3 spaces (50 - 47 = 3). Since the spinner shows numbers 1 through 10, the only way to advance exactly 3 spaces is to spin a 3. There's only one section with the number 3 out of 10 total sections.

The probability is therefore $$\frac{\text{favorable outcomes}}{\text{total possible outcomes}} = \frac{1}{10}$$, making C correct.

Let's examine why the other answers are wrong. Answer A ($$\frac{3}{10}$$) incorrectly assumes that spinning 1, 2, or 3 would all work, but spinning 1 lands you on space 48 and spinning 2 lands you on space 49 – neither wins the game. Answer B ($$\frac{2}{10}$$) might come from miscounting favorable outcomes or confusing this with a different scenario. Answer D ($$\frac{4}{10}$$) significantly overestimates the favorable outcomes, perhaps by misunderstanding the winning condition.

Strategy tip: In "exact landing" probability problems, always calculate the precise difference between your current position and target position first. Only outcomes that produce that exact difference count as favorable – getting close doesn't count in these scenarios.

There are 8 × 8 = 64 total possible outcomes when spinning twice. To get a sum of 10, we need pairs where the two numbers add to 10: (2,8), (3,7), (4,6), (5,5), and (6,4). That's 5 favorable outcomes. Therefore, P(sum = 10) = 5/64.

This is a conditional probability problem, where you need to find the probability of one event given that another event has already occurred. The key phrase "given that the card drawn is red" tells you that you're working with a restricted sample space.

In a standard deck, there are 26 red cards (13 hearts and 13 diamonds) and 26 black cards. Since you know the card is red, you're only considering those 26 red cards as your possible outcomes. Among these 26 red cards, exactly 13 are hearts.

So the probability that a red card is a heart = $$\frac{\text{number of hearts}}{\text{number of red cards}} = \frac{13}{26}$$, which is answer choice D.

Let's examine why the other answers are incorrect. Choice A ($$\frac{13}{52}$$) represents the probability of drawing a heart from the entire deck without any given information - this ignores the condition that the card is red. Choice B ($$\frac{39}{52}$$) would represent the probability of drawing a non-heart from the entire deck, which isn't what we're looking for. Choice C ($$\frac{26}{52}$$) gives the probability of drawing any red card from the entire deck, again ignoring the given condition.

When you see "given that" in a probability question, remember to adjust your sample space. You're no longer working with all possible outcomes, but only with the outcomes that satisfy the given condition. This makes your denominator smaller and changes the probability calculation.

When you encounter probability questions about "exactly" a certain number of outcomes, you need to carefully count all the ways that specific condition can be met while avoiding other possibilities.

To find the probability that exactly two rolls show the same number, let's think systematically. This means two dice show one number, and the third die shows a different number. First, choose which number appears twice (6 choices), then choose which two positions out of three show that number (3 ways), then choose what different number the third die shows (5 remaining choices). This gives us $$6 \times 3 \times 5 = 90$$ favorable outcomes.

The total possible outcomes when rolling three dice is $$6^3 = 216$$, so our probability is $$\frac{90}{216}$$, which is answer D.

Let's examine why the other answers are incorrect. Answer A ($$\frac{75}{216}$$) likely comes from miscounting the arrangements or forgetting that the third die must show a different number. Answer B ($$\frac{120}{216}$$) might result from incorrectly calculating arrangements or double-counting some cases. Answer C ($$\frac{105}{216}$$) could come from adding cases incorrectly or not properly distinguishing between "exactly two the same" versus other similar conditions.

For probability questions involving "exactly" conditions, always break the problem into clear steps: identify what you're counting, determine how many ways it can happen, calculate total possible outcomes, then form your fraction. Double-check by considering whether related cases (like all three the same, or all different) would give reasonable probabilities that add up appropriately.

This question tests developing a uniform probability model by assigning equal probability 1/n to each of n equally likely outcomes, calculating event probabilities as favorable/total. In a uniform model, outcomes are equally likely, like on a fair die where each number has P=1/6; assign equal probability P=1/n to each, event probability is count of favorable divided by total, such as even on die {2,4,6} count 3, total 6, P=3/6=1/2, and verify probabilities sum to 1. For example, fair die {1,2,3,4,5,6}, each P=1/6, event 'even'={2,4,6}, P(even)=3/6=1/2. The correct model here treats each of the 8 sections as equally likely, so P(7)=1/8 since only one section is 7. A common error is using the wrong total, like 7 instead of 8, leading to 1/7, or confusing with complementary probability like 7/8. To create the model: (1) identify equally likely outcomes (fair spinner yes), (2) count n=8, (3) assign each P=1/8, (4) verify sum=1. To calculate the event: (1) identify favorable {7}, (2) count 1, (3) divide by 8, (4) simplify to 1/8; mistakes include assuming uniform without justification or not summing to 1.

When you encounter a probability question with conditions like "given that" or "at least one," you're dealing with conditional probability. This means you need to narrow your focus to only the outcomes that meet the given condition.

First, let's identify all the ways two dice can sum to more than 9. The possible sums are 10, 11, and 12. List the outcomes: (4,6), (5,5), (6,4) for sum = 10; (5,6), (6,5) for sum = 11; and (6,6) for sum = 12. That's 6 total outcomes where the sum exceeds 9.

Now, among these 6 outcomes, count how many have at least one die showing 6: (4,6), (6,4), (5,6), (6,5), and (6,6). That's 5 outcomes with at least one 6.

The probability is $$\frac{5}{6}$$, which is answer C.

Let's see why the other answers are wrong. Answer A ($$\frac{3}{6}$$) might come from only counting outcomes where exactly one die shows 6, forgetting to include (6,6). Answer B ($$\frac{4}{6}$$) could result from excluding the (6,6) case or miscounting the favorable outcomes. Answer D ($$\frac{2}{6}$$) is far too small and might come from only counting the sum = 11 cases.

Study tip: For conditional probability problems, always work in two steps: first find all outcomes that satisfy the given condition, then count how many of those satisfy what you're looking for. Don't get distracted by the total number of possible dice outcomes (36) — focus only on your restricted sample space.

This question tests developing a uniform probability model by assigning equal probability 1/n to each of n equally likely outcomes, calculating event probabilities as favorable/total. In a uniform model, outcomes are equally likely, like on a fair die where each number has P=1/6; assign equal probability P=1/n to each, event probability is count of favorable divided by total, such as even on die {2,4,6} count 3, total 6, P=3/6=1/2, and verify probabilities sum to 1. For example, fair die {1,2,3,4,5,6}, each P=1/6, event 'even'={2,4,6}, P(even)=3/6=1/2. The correct model treats each of the 6 sections equally, with 3 labeled C, so P(C)=3/6. A common error is assigning probability per label instead of per section, like 1/3 for three labels, or treating as one outcome per letter leading to 1/6. To create the model: (1) identify equally likely outcomes (fair spinner sections yes), (2) count n=6, (3) assign each P=1/6, (4) verify sum=1. To calculate the event: (1) identify favorable C sections, (2) count 3, (3) divide by 6, (4) simplify to 1/2 but option is 3/6; mistakes include non-equal probabilities or probability >1.

This question tests developing a uniform probability model by assigning equal probability 1/n to each of n equally likely outcomes, calculating event probabilities as favorable/total. In a uniform model, outcomes are equally likely, like a fair die where each number has P=1/6; assign equal probability P=1/n to each, event probability is count of favorable divided by total, such as even on die {2,4,6} count 3, total 6, P=3/6=1/2, and verify probabilities sum to 1. For example, fair die {1,2,3,4,5,6}, each P=1/6, event 'even'={2,4,6}, P(even)=3/6=1/2. Here, the correct model assigns P=1/12 to each ticket 1-12, so P(multiple of 3)={3,6,9,12} which is 4/12. A common error is missing multiples like 12 or using wrong total, leading to incomplete sample space. To create the model: (1) identify equally likely outcomes (identical tickets yes), (2) count n=12, (3) assign each P=1/12, (4) verify sum=1. To calculate: (1) identify favorable multiples of 3, (2) count 4, (3) divide by 12, (4) simplify to 1/3 but 4/12 is equivalent; mistakes include wrong identification or not summing to 1.

This question tests developing a uniform probability model by assigning equal probability 1/n to each of n equally likely outcomes, calculating event probabilities as favorable/total. In a uniform model, outcomes are equally likely, like a fair die where each number has P=1/6; assign equal probability P=1/n to each, event probability: count favorable, divide by total (even on die: {2,4,6} count 3, total 6, P=3/6=1/2), verify sum to 1. For example, fair die {1,2,3,4,5,6}, each P=1/6, event 'even'={2,4,6}, P(even)=3/6=1/2. Here, the correct model for MATH has slips {M,A,T,H}, each P=1/4, P(A)=1/4. A common error is treating repeated letters as one or using word length incorrectly like 1/3. To create the model: (1) identify equally likely slips (mixed yes), (2) count n=4, (3) assign each P=1/4, (4) verify sum=1. To calculate the event: (1) identify favorable {A}, (2) count 1, (3) divide by 4, (4) simplify to 1/4; mistakes include assuming non-uniform or wrong total.